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The equation \(x^{\frac{3}{4}(\text{log}_2\,x)^2+(\text{log}_2\,x)-\frac{5}{4}}\) = √2 has

(a) at least one real solution 

(b) exactly one irrational solution 

(c) exactly three real solutions 

(d) all of the above.

1 Answer

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Best answer

(c) exactly three real solutions 

Given, \(x^{\frac{3}{4}(\text{log}_2\,x)^2}\) \(+\text{log}_2\,x-\frac{5}{4}\) = √2

Taking log to the base 2 of both the sides, we have

\(\bigg[\frac{3}{4}^{(\text{log}_2\,x)^2+(\text{log}_2)-\frac{5}{4}}​​\bigg]\) log2 x = log2 √2 = log2 \(2^{\frac{1}{2}}\) = \(\frac{1}{2}\)log2 = \(\frac{1}{2}\)

Let us assume log2x = a. Then,

\(\bigg(\frac{3}{4}a^2+a-\frac{5}{4}\bigg)a\) = \(\frac{1}{2}\) ⇒ 3a3 + 4a2 - 5a = 2

⇒ 3a3 + 4a2 – 5a – 2 = 0. 

Using hit and trial method check for a = 1. 

f(a) = 3a3 + 4a2 – 5a – 2 ⇒ f(1) = 3.13 + 4.12 – 5.1 – 2 = 0 

∴ (a – 1) is a factor of 3a3 + 4a2 – 5a – 2

∴ Now by dividing 3a3 + 4a2 – 5a – 2 by (a – 1), we get 

3a3 + 4a2 – 5a – 2 = (a – 1) (3a + 1) (a + 2) = 0

⇒ a = 1 or a = \(-\frac{1}{3}\) or a = - 2

⇒ log2x = 1 or log2x = \(-\frac{1}{3}\) or log2x = - 2

⇒ x = 21 = 2 or x = 2-1/3 or x = 2-2\(\frac{1}{4}\)

∴ The given equation has exactly three real solutions, wherein x = 2–1/3 is irrational.

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